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Question
Calculate the total torque acting on the body shown in the following figure about the point O.
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Solution
Torque about a point = Total force × Perpendicular distance
Let the anticlockwise torque and clockwise acting torque be positive and negative, respectively.
Torque at O due to 5 N force is zero as it is passing through O.
Torque at O due to 15 N force,
\[\tau_1 = 15 \times 6 \times {10}^{- 2} \times \sin37^\circ\]
\[ = 15 \times 6 \times {10}^{- 2} \times \frac{3}{5}\]
\[ = 0 . 54 N - m .........\left(\text{anticlockwise} \right)\]
Torque at O due to 10 N force,
\[\tau_2 = 10 \times 4 \times {10}^{- 2} = 0 . 4 N - m ........\left(\text{clockwise} \right)\]
Torque at O due to 20 N force,
\[\tau_3 = 20 \times 4 \times {10}^{- 2} \times \sin30^\circ\]
\[ = 20 \times 4 \times {10}^{- 2} \times \frac{1}{2}\]
\[ = 0 . 4 N - m .........\left(\text{Anticlockwise} \right)\]
Resultant torque acting at O,
\[\tau = \left( 0 . 54 - 0 . 4 + 0 . 4 \right)\]
\[ = 0 . 54 N - m \left(\text{Anticlockwise} \right)\]
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